If G is the geometric mean of P(A), P(B), P(C) and P(D), then what is 9G equal to ?

For the next three (03) items that follow : Let A, B, C and D be mutually exclusive and exhaustive events and \( \frac{P(A)}{2} = \frac{P(B)}{3} = \frac{P(C)}{5} = \frac{P(D)}{8} \).

  1. A. 17^{\frac{1}{4}}
  2. B. 15^{\frac{1}{4}} ✓
  3. C. 13^{\frac{1}{4}}
  4. D. 11^{\frac{1}{4}}

Correct Answer: B. 15^{\frac{1}{4}}

Explanation

G = (P(A)P(B)P(C)P(D))^(1/4) = ((2/18)*(3/18)*(5/18)*(8/18))^(1/4) = (240 / 18^4)^(1/4). Since 240 = 16 * 15, G = (2/18) * 15^(1/4) = (1/9) * 15^(1/4). Thus, 9G = 15^(1/4).

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